Even more intergration problems! (1 Viewer)

[CrashOveride]

Ok i think i got it.

This will sound pathetic, but here goes

I'm applying the abovementioned limits to a certain integral....so writing the integral and putting the brackets around it..then the limits on those is the same as writing the limits above and below the integral sign, its the same thing.....i was previously assuming this certain integral to identify with some U part but i could only do that if it had the same limits..... i hope i made sense

 

[CM_Tutor]

Originally posted by CrashOveride
Actually just for reference, could you write it up

I may have more questions to follow

No problem with more questions, and here's the full working:

Theorem: If U_n = ∫ (-1 --> 0) xⁿ(1 + x)^1/2 dx, for integers n ≥ 0, then U_n = -2n * U_n-1 / (2n + 3), for n ≥ 1

Proof: U_n = ∫ (-1 --> 0) xⁿ(1 + x)^1/2 dx, for integers n ≥ 0
= ∫ (-1 --> 0) xⁿ d[(1 + x)^3/2 / (3 / 2)]
= [2xⁿ(1 + x)^3/2 / 3] (-1 --> 0) - ∫ (-1 --> 0) 2(1 + x)^3/2 / 3 dxⁿ
= (2 / 3) * [0ⁿ1^3/2 - (-1)ⁿ0^3/2] - (2n / 3) * ∫ (-1 --> 0) x^n-1(1 + x)^1+1/2 dx, for integers n ≥ 1
= (2 / 3) * [0 - 0] - (2n / 3) * ∫ (-1 --> 0) x^n-1(1 + x)(1 + x)^1/2 dx
= (-2n / 3) ∫ (-1 --> 0) x^n-1(1 + x)^1/2 + xⁿ(1 + x)^1/2 dx
= (-2n / 3) ∫ (-1 --> 0) x^n-1(1 + x)^1/2 dx + (-2n / 3) ∫ (-1 --> 0) xⁿ(1 + x)^1/2 dx
= (-2n / 3) * (U_n-1 + U_n)

So, U_n(1 + 2n / 3) = (-2n / 3) * U_n-1
U_n(3 + 2n) / 3 = -2n * U_n-1 / 3
So, U_n = -2n * U_n-1 / (2n + 3), for integers n ≥ 1, as required.

 

[CrashOveride]

Thanks CM.

One more:

If I_n = ∫e^mx tanⁿx dx

Show I_n = {[e^mx tan^n-1x] / [n-1] } - [m / (n-1)].I_n-1 - I_n-2

I get pretty close with: I_n = { [e^mx . tan^n-1x ] / n } - (n/m)I_n-1 - I_n-2

 

[CM_Tutor]

Originally posted by CrashOveride
If I_n = ∫e^mx tanⁿx dx

Show I_n = {[e^mx tan^n-1x] / [n-1] } - [m / (n-1)].I_n-1 - I_n-2

I get pretty close with: I_n = { [e^mx . tan^n-1x ] / n } - (n/m)I_n-1 - I_n-2

I_n = ∫ e^mxtanⁿx dx, for integers n ≥ 0
= ∫ tanⁿx d(e^mx / m)
= tanⁿx * e^mx / m - ∫ e^mx / m dtanⁿx
= e^mxtanⁿx / m - (1 / m) ∫ e^mx * ntan^n-1x * sec²x dx, n ≥ 1
= e^mxtanⁿx / m - (n / m) ∫ e^mxtan^n-1x * (1 + tan²x) dx, noting that 1 + tan²x = sec²x
= e^mxtanⁿx / m - (n / m) ∫ e^mxtan^n-1x + e^mxtanⁿ⁺¹x dx
= e^mxtanⁿx / m - (n / m) * (I_n-1 + I_n+1)

So, (n / m) * I_n+1 = e^mxtanⁿx / m - (n / m) * I_n-1 - I_n
I_n+1 = (m / n) * e^mxtanⁿx / m - (m / n) * (n / m) * I_n-1 - (m / n) * I_n
I_n+1 = e^mxtanⁿx / n - I_n-1 - (m / n) * I_n
I_n+1 = e^mxtanⁿx / n - (m / n) * I_n - I_n-1

Now, replace n + 1 by n, and we get, for integers n ≥ 2:

So, I_n = e^mxtan^n-1x / (n - 1) - [m / (n - 1)] * I_n-1 - I_n-2, for integers n ≥ 2, as required.

 

[CrashOveride]

Thanks CM. I got up to your 7th line except insttead of making I_n+1 the subject and going from there I replaced the original I_n with (n-1) and then re-arranged for I_n.

It gave me a slightly different answer, but this would still be correct for other purposes?

Just that we had to actually show a certain thing in this question.

And also ive noticed we can never allow an 'n' such that one of our 'I' will have a negative number? It just doenst work for n<0 ?

 

[CM_Tutor]

Originally posted by CrashOveride
Thanks CM. I got up to your 7th line except insttead of making I_n+1 the subject and going from there I replaced the original I_n with (n-1) and then re-arranged for I_n.

It gave me a slightly different answer, but this would still be correct for other purposes?

Just that we had to actually show a certain thing in this question.

Did you replace all the n's by (n - 1)'s? If you did, it should be OK, but it should also have given the same answer. If you only replaced some of them, then you have not treated both sides of the equation in the same way, and in that case your answer is not valid.

And also ive noticed we can never allow an 'n' such that one of our 'I' will have a negative number? It just doenst work for n<0 ?

It might - it depends on the question. In this case, it won't, as n is being used to mean tanⁿx = (tan x)ⁿ. In the case of n = -1 in the original gives the integral &int; e^mxtan^-1x dx, which is not at all the same thing as
&int; e^mx(tan x)^-1 dx = &int; e^mx / tan x dx = &int; e^mxcot x dx, is it?

Furthermore, if you read my integration by parts working carefully, you may note that I change the domain of n in the working at the point where the problem arises. In the above (and indeed, in most cases), it occurs when the differentiation is performed. Here, I have rewritten dtanⁿx as ntan^n-1x * sec²x dx, which I can only do if n is a positive integer.

 

[CrashOveride]

Originally posted by CM_Tutor
Did you replace all the n's by (n - 1)'s? If you did, it should be OK, but it should also have given the same answer. If you only replaced some of them, then you have not treated both sides of the equation in the same way, and in that case your answer is not valid.

Right you are again, yeah forgot an 'n'. Fixed now. Thanks CM